Problem Set 4
.±
 Solutions
J. Scholtz
October 23, 2006
Question 1: 2.5/8
We can pretty much follow the proof of theorem 2.23:
Let
T
:
V
→
W
and
β
,
β
0
be the ordered bases for
V
and similarly with
γ
’s for
W
. Then we can consider the transformation: [
T
]
γ
β
Q
= [
T
]
γ
β
[
I
]
β
β
0
= [
T
]
γ
β
0
We can
also check that
P

1
switches
γ
to
γ
0
:
PP

1
=
I
γ
, hence the result is in
γ
basis
but the argument of
P
is in the
γ
0
basis. Moreover:
PP

1
= [
I
]
γ
γ
0
P

1
=
I
γ
,
therefore
P

1
= [
I
]
γ
0
γ
. And thence by repeating the argument from ﬁrst part:
[
T
]
γ
0
β
0
=
P

1
[
T
]
γ
β
Q
Question 2: 2.5/10
If
A
and
B
are similar matrices, then there exists an invertible matrix
Q
, such
that
B
=
Q

1
AQ
. Then by exercise 13 in section 2.3: Tr(
CD
) = Tr(
DC
),
hence:
Tr(
B
) = Tr(
Q

1
AQ
) = Tr(
AQQ

1
) = Tr(
AI
) = Tr(
A
)
Question 3: 2.6/4
Let
V
=
R
3
, and let there be
f
1
,f
2
,f
3
∈
V
*
such that:
f
1
=
x

2
y
f
2
=
x
+
y
+
z
f
3
=
y

3
z
Since we know that dim(
V
*
) = dim(
V
) = 3, since
V
is ﬁnite dimensional, then it
is enough to show span or linear independence. Let us show linear independence.
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 Spring '08
 GUREVITCH

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